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I am trying to render a 3D model using OpenGL. For the projection and transformation matrices, I am using glm. I've got my model on the screen and it works just like I intended it to; except one small problem.

I am setting the model's translation matrix as

glm::translate(glm::vec3(0, 0, 4))

to move the model a little bit forward so that I can see it. Since in OpenGL, by default, negative z is out towards the 'camera' and positive z is forward, I expected this to work but it doesn't. It only works if I set it to

glm::translate(glm::vec3(0, 0, -4))

But this seems weird to me, as I am setting my zNear to 0.01 and zFar to 1000. Is glm's z values flipped or am I doing something wrong here?

Here is my code: 

glm::mat4 rotation = glm::mat4(1.0f);
glm::mat4 translation = glm::translate(glm::vec3(0, 0, -4));
glm::mat4 scale = glm::mat4(1.0f);

glm::mat4 modelMatrix = translation * rotation * scale;

glm::mat4 projectionMatrix = glm::perspective(70.0f, aspectRatio, 0.01f, 1000.0f);

glm::mat4 transformationMatrix = projectionMatrix * modelMatrix;
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1 Answer 1

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Since in OpenGL, by default, negative z is out towards the 'camera' and positive z is forward

You're wrong. OpenGL uses a right-handed coordinate-system where x is right, y is up and z is into the camera, therefore negative z is forwards.

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  • \$\begingroup\$ Oh, I always assumed the opposite. So setting zNear to 0.01 and zFar to 1000 in a perspective matrix is actually setting them to -0.01 and -1000 respectively? \$\endgroup\$
    – Mertcan Ekiz
    Jul 31, 2015 at 2:08
  • \$\begingroup\$ Well thats hard to explain. They are not stored directly, they are used to calculate the projection-matrix. But in some sense you could say that you'd see -0.1 to -1000.0, yes. \$\endgroup\$
    – tkausl
    Jul 31, 2015 at 2:12

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